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Particular integral of sin(ax+b)

  1. Oct 24, 2012 #1
    I can prove that the particular integral of sin(ax+b) is given by

    [itex]\frac{1}{f(-a^{2})} sin(ax+b)[/itex]

    That's not an issue but what happens if f(-a2) is 0 ?

    I see that it is given by

    [itex] x \frac{1}{f'(-a^{2})} sin(ax+b)[/itex]

    but I can't prove this though I know to prove this for eax


    Thanks a lot :)
     
  2. jcsd
  3. Oct 24, 2012 #2

    HallsofIvy

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    Staff Emeritus
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    I don't understand what you are saying. The general anti-derivative of sin(ax+ b) is -(1/a) cos(ax+ b)+ C for any constant C. Yes, you can take C= 0 but where did "f" come from?
     
  4. Oct 24, 2012 #3
    [itex]
    f(D)=(D^{n}+k_{1}D^{n-1}+k_{2}D^{n-2}+...+k_{n-1}D+k_{n})[/itex]

    and I mean the differential equation is given by

    [itex]
    (D^{n}+k_{1}D^{n-1}+k_{2}D^{n-2}+...+k_{n-1}D+k_{n})y=X
    \\
    or \\f(D)y=X

    [/itex]

    Thanks a lot :)
     
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